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Form a right triangle with sides L1, L2 and an hypotenue 2R. According to the Pytagorean theorem, L1^2 + L2^2 = 4R^2
using the cosines law we can obtain L1 and L2 in terms of R and an angle theta.
L1^2=R^2+R^2-2*R*R*Cos[theta]
L2^2=R^2+R^2-2*R*R*Cos[180-theta]
So, the sum of L1^2 + L2^2 should be 4*R^2, therefore,
4*R^2=R^2+R^2-2*R*R*Cos[theta]+R^2+R^2-2*R*R*Cos[180-theta].
Any terms of R goes and it results in a trigonometric identity: Cos[180-theta]=-Cos[theta]
With this identity, we replace one of these terms in the previous ecuations of L1^2 or L2^2 and do the sum again with the new term:
4*R^2=R^2+R^2+2*R*R*Cos[theta]+R^2+R^2-2*R*R*Cos[theta]
And finally it's proven, 4*R^2=4*R^2, thus the circle does intersect the vertex joining L1 and L2 in any right triangle. |
